cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A250001 Number of arrangements of n circles in the affine plane.

Original entry on oeis.org

1, 1, 3, 14, 173, 16951
Offset: 0

Views

Author

Jon Wild, May 16 2014

Keywords

Comments

Two circles are either disjoint or meet in two points. Tangential contacts are not allowed. A point belongs to exactly one or two circles. Three circles may not meet at a point. The circles may have different radii.
This is in the affine plane, rather than the projective plane.
Two arrangements are considered the same if one can be continuously changed to the other while keeping all circles circular (although the radii may be continuously changed), without changing the multiplicity of intersection points, and without a circle passing through an intersection point. Turning the whole configuration over is allowed.
Several variations are possible:
- straight lines instead of circles (see A241600).
- straight lines in general position (see A090338).
- curved lines in general position (see A090339).
- allow circles to meet tangentially but without multiple intersection points (begins 1, 5, ...); more terms are needed.
- again use circles, but allow multiple intersection points (also begins 1, 5, ...); more terms are needed.
- use ellipses rather than circles.
- a question from Walter D. Wallis: what if the circles must all have the same radius?
a(1)-a(5) computed by Jon Wild.
a(n) >= A000081(n+1) - Benoit Jubin, Dec 21 2014. More precisely, there are A000081(n+1) ways to arrange n circles if no two of them meet. - N. J. A. Sloane, May 16 2017
From Daniel Forgues, Aug 08-09 2015: (Start)
A representation for the diagrams in a250001.jpg (in the same order):
a(1) = 1: {{2}};
a(2) = 3: {{2, 3}, {2, 4}, {4, 6}};
a(3) = 14: {{2, 4, 8}, {2, 3, 6}, {2, 3, 4}, {2, 3, 5}, {4, 6, 5},
{4, 6, 15}, {2, 6, 9}, {4, 6, 12}, {2, 8, 12}, {30, 42, 70},
{?, ?, ?}, {?, ?, ?}, {15, 21, 35}, {?, ?, ?}}.
In lexicographic order:
a(3) = 14: {{2, 3, 4}, {2, 3, 5}, {2, 3, 6}, {2, 4, 8}, {2, 6, 9},
{2, 8, 12}, {4, 6, 5}, {4, 6, 12}, {4, 6, 15}, {15, 21, 35},
{30, 42, 70}, {?, ?, ?}, {?, ?, ?}, {?, ?, ?}}.
The smallest integers greater than 1 are used for the representation:
(p_1)^(a_1)*...*(p_m)^(a_m), where
0 <= a_i <= n, for 1 <= i <= m;
(a_1)+...+(a_m) > 0.
Could the Venn diagram interpretation (of the k-wise, 1 <= k <= n, common divisors of k numbers from each subset) reveal a pattern?
Does this representation work for more complex diagrams? (End)
Once you get to n=5, geometric concerns mean that not all topologically-conceivable arrangements are actually circle-drawable. My program enumerated 16977 conceivable arrangements of 5 pseudo-circles, and Christopher Jones and I together have figured out how to show that 26 of these arrangements are not actually circle-drawable. So it seems that a(5) = 16951. This entry will be updated soon, and there will be a new sequence for the number of topologically-conceivable arrangements. - Jon Wild, Aug 25 2016 [The counts in this comment were amended by Jon Wild on Aug 30 2016. I apologize for taking so long to make the corrections here. - N. J. A. Sloane, Jun 11 2017]
a(n) <= 7*13^(binomial(n,3) + binomial(n,2) + 3n - 1) is a (loose) upper bound, see Reddit link. I believe XkF21WNJ's reply shaves off a factor of 13^3 from this bound for all n > 1. - Linus Hamilton, Apr 14 2019
A good upper bound for a(6) is given in sequence A288559, which counts the arrangements of pseudo-circles, i.e. the topologically conceivable arrangements mentioned above, which are not all necessarily realizable with true circles. The number of arrangements of 6 pseudo-circles was found by Andrii Shportko and Jon Wild to be 17,552,169. - Jon Wild, Jun 03 2025
In A288559, a(5) included 26 non-circularizable pseudocircle arrangements, which generated in turn 132,546 6-pseudocircle descendants. These descendants must be excluded from A250001, which means that a tighter upper bound for A250001(6) is 17,419,623. - Andrii Shportko, Jun 06 2025

Examples

			a(2) = 3, because two circles can either be next to each other, overlap with two intersection points, or one may be located within the other (of larger radius). (As per the first comment, the limiting case where they touch in one point is [somewhat arbitrarily] excluded. This would add two more independent configurations, where one touched the other "from inside" or "from outside".) - _M. F. Hasler_, May 03 2025
		

References

  • Jon Wild, Posting to Sequence Fans Mailing List, May 15 2014.

Crossrefs

Row sums of A261070.
Apart from first term, row sums of triangles A249752, A252158, A285996, A274776, A274777.
See A275923 and A275924 for the connected arrangements. See also A288554-A288568.
Cf. A132101 (one-dimensional analog).

Extensions

a(4) is 173, not 168. Corrected by Jon Wild, Aug 08 2015
A duplicate pair of configurations in an older file was spotted by Manfred Scheucher, Aug 13 2016. The pdf and svg files here are now correct.

A249752 Triangle read by rows: T(n,k) = number of arrangements of n circles in the affine plane having k outer circles.

Original entry on oeis.org

1, 1, 2, 3, 5, 6, 14, 52, 60, 47
Offset: 1

Views

Author

Omar E. Pol, Dec 14 2014

Keywords

Comments

Consider the arrangements of n circles described in A250001.
T(n,k) is the number of arrangements of n circles in the affine plane whose perimeters are formed with parts from k circles. - Omar E. Pol, Aug 09 2015
From Omar E. Pol, Jul 06 2016: (Start)
Observation 1: column 1 gives A250001, at least if 1<=n<=4.
Observation 2: sum of n-th row = T(n+1,1), at least if 1<=n<=3. (End)

Examples

			Triangle begins:
1;
1,   2;
3,   5,  6;
14, 52, 60, 47;
...
		

Crossrefs

Sum of n-th row = A250001(n).
Leading diagonal gives A274702.

Extensions

Corrected a(5)-a(6) and added a(7)-a(10) from Omar E. Pol, Oct 13 2015

A261070 Irregular triangle read by rows: T(n,k) is the number of arrangements of n circles with 2k intersections (using the same rules as A250001).

Original entry on oeis.org

1, 1, 2, 1, 4, 4, 2, 4, 9, 15, 15, 31, 24, 35, 44, 20, 50
Offset: 0

Views

Author

Benoit Jubin, Aug 08 2015

Keywords

Comments

Length of n-th row: 1 + (n-1)n/2 (for a configuration for T(n,(n-1)n/2), consider n circles of radius 1 and centers at (k/n,0) for 1<=k<=n).
The generating function down the column k=1 is 1+z^2 *C^2(z) *[C^2(z)+C(z^2)]/ (2*[1-z*C(z)]) = 1+ z^2 +4*z^3 +15*z^4+ 50*z^5+...where C(z) = 1+z+2*z^2+4*z^3+... is the g.f. of A000081 divided by z; eq. (78) in arXiv:1603.00077. - R. J. Mathar, Mar 05 2016

Examples

			n\k 0  1  2  3  4  5  6
0   1
1   1
2   2  1
3   4  4  2  4
4   9 15 15 31 24 35 44
5  20 50  .  .  .  .  .  .  .  .  .
		

Crossrefs

Row sums give A250001.
Cf. A000081, A152947, A249752, A252158, A280786 (column k=1)

Formula

A250001(n) = Sum_{k>=0} T(n,k).
A000081(n+1) = T(n,0).

Extensions

T(4,2)..T(5,0) (6 terms) from Travis Vasquez, Nov 28 2024

A274776 Irregular triangle read by rows: T(n,k) = number of arrangements of n circles in the affine plane forming k regions, including the regions that do not belong to the circles.

Original entry on oeis.org

1, 0, 2, 1, 0, 0, 4, 4, 2, 0, 4, 0, 0, 0
Offset: 1

Views

Author

Omar E. Pol, Jul 06 2016

Keywords

Comments

Consider the arrangements of n circles described in A250001.
Note that the sum of the 4th row must be equal to A250001(4) = 173.

Examples

			Triangle begins:
1;
0, 2, 1;
0, 0, 4, 4, 2, 0, 4;
0, 0, 0, ...
...
For n = 3 and k = 5 there are 2 arrangements of 3 circles in the affine plane forming 5 regions, including the regions that do not belong to the circles, so T(3,5) = 2.
For n = 3 and k = 6 there are no arrangements of 3 circles in the affine plane forming 6 regions, including the regions that do not belong to the circles, so T(3,6) = 0.
Of course, there is a right triangle of all zeros starting from the second row.
		

Crossrefs

Sum of n-th row = A250001(n).
First differs from A274777 at a(10).

Formula

T(n,k) = A274818(n,k)/k.

A274777 Irregular triangle read by rows: T(n,k) = number of arrangements of n circles in the affine plane forming k regions, excluding the regions that do not belong to the circles.

Original entry on oeis.org

1, 0, 2, 1, 0, 0, 4, 4, 2, 1, 3, 0, 0, 0
Offset: 1

Views

Author

Omar E. Pol, Jul 06 2016

Keywords

Comments

In other words: not counting the regions between circles.
Consider the arrangements of n circles described in A250001.
Note that the sum of the 4th row must be equal to A250001(4) = 173.

Examples

			Triangle begins:
1;
0, 2, 1;
0, 0, 4, 4, 2, 1, 3;
0, 0, 0, ...
...
For n = 3 and k = 5 there are 2 arrangements of 3 circles in the affine plane forming 5 regions, excluding the regions that do not belong to the circles, so T(3,5) = 2.
For n = 3 and k = 6 there is only one arrangement of 3 circles in the affine plane forming 6 regions, excluding the regions that do not belong to the circles, so T(3,6) = 1.
Of course, there is a right triangle of all zeros starting from the second row.
		

Crossrefs

Sum of n-th row = A250001(n).
First differs from A274776 at a(10).

Formula

T(n,k) = A274819(n,k)/k.

A285996 Triangle read by rows, 1<=k<=n, T(n,k) = number of arrangements of n circles in the affine plane having k separated islands.

Original entry on oeis.org

1, 2, 1, 11, 2, 1, 156, 14, 2, 1
Offset: 1

Views

Author

Omar E. Pol, May 21 2017

Keywords

Comments

Consider the rules for the arrangements of n circles described in A250001.
Note that T(4,1) = 156 includes the arrangement of 4 circles in which there is a little circle that is surrounded by the union of three solid circles, because the little circle is an inland island, or lake island, which does not count. So there is only one separated island. Hence T(4,2) = 14 does not include the mentioned arrangement.
Question: is 1 together with the first column of the triangle the same as A275923?

Examples

			Triangle begins:
1;
2,   1;
11,  2,  1;
156, 14, 2, 1;
		

Crossrefs

Another version of A252158.
Row sums give A250001, n >= 1.
Right border gives A000012.

A287149 Number of arrangements of n circles in the affine plane having only one island or region, which is formed by the union of all solid circles of the arrangement.

Original entry on oeis.org

1, 1, 2, 11, 155
Offset: 0

Views

Author

Omar E. Pol, May 21 2017

Keywords

Comments

A subset of the arrangements of n circles described in A250001.
Note that a(4) = 155 does not include the arrangement of four circles in which there is a little circle that is surrounded by the union of three solid circles, because in that arrangement there are two islands or regions, not one.

Crossrefs

1 together with the first column of A252158.
First differs from A275923 at a(4).
Cf. A250001.
Showing 1-7 of 7 results.